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3 years ago

Integer or simplified fraction:)! help pls<3

Mathematics

1 answer:

Leno4ka [110]3 years ago

4 0

Answer:

$\dfrac{f^0}{j^{-3}} =8$

Step-by-step explanation:

The question relates to the evaluation of an expression given the values of the variables of the expression;

The given expression is $\dfrac{f^0}{j^{-3}}$

The values of the terms at which the given expression is to be evaluated are;

f = -4 and j = 2

By plugging in the values, we get;

At f = -4 and j = 2, $\dfrac{f^0}{j^{-3}} = \dfrac{(-4)^0}{2^{-3}}$

(-4)⁰ = 1

2⁻³ = 1/8

We get;

$\dfrac{f^0}{j^{-3}} = \dfrac{(-4)^0}{2^{-3}} = \dfrac{1}{\left (\dfrac{1}{8} \right )} =8$

Therefore;

$For \ f = -4 \ and \ j = 2, \ \dfrac{f^0}{j^{-3}} = 8.$

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Kristy dad makes 3 times as much per hour as she does. If he makes $48 per hour, how much per hour, how much per hour does krist

Jlenok [28]

Answer:

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3 years ago

Consider the line integral Z C (sin x dx + cos y dy), where C consists of the top part of the circle x 2 + y 2 = 1 from (1, 0) t

pashok25 [27]

Direct computation:

Parameterize the top part of the circle $x^2+y^2=1$ by

$\vec r(t)=(x(t),y(t))=(\cos t,\sin t)$

with $0\le t\le\pi$ , and the line segment by

$\vec s(t)=(1-t)(-1,0)+t(2,-\pi)=(3t-1,-\pi t)$

with $0\le t\le1$ . Then

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)$

$=\displaystyle\int_0^\pi(-\sin t\sin(\cos t)+\cos t\cos(\sin t)\,\mathrm dt+\int_0^1(3\sin(3t-1)-\pi\cos(-\pi t))\,\mathrm dt$

$=0+(\cos1-\cos2)=\boxed{\cos1-\cos2}$

Using the fundamental theorem of calculus:

The integral can be written as

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=\int_C\underbrace{(\sin x,\cos y)}_{\vec F}\cdot\underbrace{(\mathrm dx,\mathrm dy)}_{\vec r}$

If there happens to be a scalar function $f$ such that $\vec F=\nabla f$ , then $\vec F$ is conservative and the integral is path-independent, so we only need to worry about the value of $f$ at the path's endpoints.